1/2log 8 base 4 + log 32 base 4 - log 2 base 4
Log base 10(x^2- 4x + 7)=2
To simplify the expression 1/2log8 base 4 + log32 base 4 - log2 base 4, we can use the logarithmic properties.
Step 1:
Using the logarithmic property log base b of a = log base b of c^x, we can rewrite 1/2log8 base 4 as log4 base 8^(1/2).
Step 2:
The square root of 8 is equal to 2^3/2, so we can rewrite log4 base 8^(1/2) as log4 base (2^3/2).
Step 3:
Using the logarithmic property log base b of a = log base b of c/d, we can rewrite log4 base (2^3/2) as log4 base 2^3 - log4 base 2.
Step 4:
Simplifying, log4 base 2^3 equals 3, and log4 base 2 equals 1, so the expression becomes 3 - 1.
Step 5:
Simplifying further, 3 - 1 equals 2.
Therefore, the simplified expression is 2.
Moving on to the next question:
To solve the equation log base 10 (x^2-4x+7) = 2, we can use the logarithmic definition.
Step 1:
Using the logarithmic definition, we can rewrite the equation as 10^2 = x^2 - 4x + 7.
Step 2:
Simplifying, 10^2 is equal to 100, so the equation becomes 100 = x^2 - 4x + 7.
Step 3:
Rearranging the equation, we get x^2 - 4x + 7 - 100 = 0.
Step 4:
Simplifying further, x^2 - 4x - 93 = 0.
Step 5:
To solve this quadratic equation, we can factor or use the quadratic formula. However, in this case, the quadratic equation cannot be factored easily, so we can use the quadratic formula.
Step 6:
Using the quadratic formula x = (-b ± √(b^2 - 4ac)) / (2a), where a = 1, b = -4, and c = -93, we find:
x = (-(-4) ± √((-4)^2 - 4(1)(-93))) / (2(1))
= (4 ± √(16 + 372)) / 2
= (4 ± √388) / 2
= (4 ± √(2^2 * 97)) / 2
= (4 ± 2√97) / 2
= 2 ± √97.
Therefore, the solutions to the equation are x = 2 + √97 and x = 2 - √97.
To solve the equations:
1) 1/2log 8 base 4 + log 32 base 4 - log 2 base 4:
We can simplify the expression using the log properties, particularly the product and quotient rules:
logb (xy) = logb(x) + logb(y) for the product rule,
and logb (x/y) = logb(x) - logb(y) for the quotient rule.
Using the above log properties, we can rewrite the equation:
1/2log 8 base 4 + log 32 base 4 - log 2 base 4
= 1/2 log4(8) + log4(32) - log4(2)
= 1/2 * 3 + 5 - 1
= 3/2 + 5 - 1
= 7/2
So, the solution is 7/2.
2) Log base 10(x^2 - 4x + 7) = 2:
To solve this equation, we need to use the logarithmic property that states:
If logb(x) = y, then b^y = x.
Using this property, we can rewrite the equation:
10^2 = x^2 - 4x + 7
Simplifying further, we have:
100 = x^2 - 4x + 7
Rearranging the equation to get it in standard quadratic form:
x^2 - 4x + 7 - 100 = 0
x^2 - 4x - 93 = 0
Now we have a quadratic equation that can be solved using factoring, completing the square, or the quadratic formula.
To solve the equation log base 10(x^2 - 4x + 7) = 2, we can rewrite it in exponential form using the definition of logarithms:
10^2 = x^2 - 4x + 7
Simplifying, we have:
100 = x^2 - 4x + 7
Rearranging the equation to the standard quadratic form, we get:
x^2 - 4x + 7 - 100 = 0
Simplifying further, we have:
x^2 - 4x - 93 = 0
To factor this quadratic equation, we need to find two numbers that multiply to give -93 and add up to -4. The numbers are -9 and 11. Therefore, the factored form of the equation is:
(x - 9)(x + 11) = 0
Setting each factor equal to zero, we have:
x - 9 = 0 or x + 11 = 0
Solving these equations, we get:
x = 9 or x = -11
Hence, the solutions to the equation are x = 9 and x = -11.